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我正在使用具有确定性用户定义函数的虚拟列（oracle 11g），该函数将行的主键作为参数并返回一个简单的标量值。虚拟列更新没有任何问题，但是当我更新表时，它会抛出错误：- ora-00054 资源繁忙并在指定 nowait 的情况下获取或在 oracle 中超时。我的表结构和功能如下：-

 -----------------------------------------------------------------------
   id   employee_name    employee_dept  employee_leaves (vir column)
 -----------------------------------------------------------------------
   2     patrick           mgmt         getEmpLeaves(id)  
   3      jack             sales            "
 -----------------------------------------------------------------------

     create or replace function getEmpLeaves(empId number) 
        return number 
          DETERMINISTIC
       is
           emp_leaves number;
       begin
           select leaves into emp_leaves from tbl_emp_leaves 
           where tbl_emp_leaves.id = empId;
           return emp_leaves;
      end ;
    -------------------------------------------------------------

如何克服这个错误？

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2 回答 2

2

我不会寻找这个错误的原因。
一个简短的回答是：从表中删除这个虚拟列，然后创建一个视图：

create view vw_employees AS
SELECT t.id, t.employee_name, t.employee_dept, x.leaves As employee_leaves
FROM tbl_employees t
JOIN tbl_emp_leaves x
ON t.id = x.id;

长答案：请看下面的简单测试用例：

create table tbl_emp_leaves as
select object_id as id, trunc(dbms_random.value(0,100)) as leaves
from all_objects;
alter table tbl_emp_leaves add primary key( id );

create or replace function getEmpLeaves(empId number) 
        return number 
          DETERMINISTIC
       is
           emp_leaves number;
       begin
           select leaves into emp_leaves from tbl_emp_leaves 
           where tbl_emp_leaves.id = empId;
           return emp_leaves;
      end ;
      /

create table tbl_employees as
select object_id as id, object_name as employee_name, object_type as employee_dept
from all_objects;

alter table tbl_employees 
add employee_leaves as ( getEmpLeaves(id)); 

create view vw_employees AS
SELECT t.id, t.employee_name, t.employee_dept, x.leaves As employee_leaves
FROM tbl_employees t
JOIN tbl_emp_leaves x
ON t.id = x.id;

现在比较两个简单查询的性能：

SQL> set timing on;
SQL> select sum(employee_leaves) from vw_employees;

SUM(EMPLOYEE_LEAVES)
--------------------
             3675425

Elapsed: 00:00:00.07
SQL> select sum(employee_leaves) from tbl_employees;

SUM(EMPLOYEE_LEAVES)
--------------------
             3675425

Elapsed: 00:00:03.09

3.09 秒与 0.07 秒 - 您会看到基于函数的虚拟列比简单连接慢 44 倍（即：4400%）。

于 2016-04-12T19:00:57.460 回答
1

当我创建一个选择另一个表的函数时，我遇到了类似的问题。

关键字DETERMINISTIC意味着函数总是为给定的输入参数返回相同的值。开发人员有责任确保这一点。

Oracle 允许这样的函数：

create or replace function getNumber(x number) 
    return number DETERMINISTIC is
begin
  return DBMS_RANDOM.RANDOM;
end;

尽管这实际上与确定性函数相反。

UPDATE tbl_emp_leaves SET leaves = ...语句后函数的返回值发生变化。

所以，我不得不同意 kordirko 的回答：删除这个虚拟列并创建一个视图或使用触发器来设置 column 的值employee_leaves。

于 2016-04-12T20:46:17.537 回答